proof of primitive element theorem

Theorem.

Let F and K be arbitrary fields, and let K be an extension of F of finite degree. Then there exists an elementα∈K such that K=F⁢(α) if and only if there are finitely many fields L with F⊆L⊆K.

Proof.

Let F and K be fields, and let [K:F]=n be finite.

Suppose first that K=F⁢(α). Since K/F is finite, α is algebraic over F. Let m⁢(x) be the minimal polynomial of α over F. Now, let L be an intermediary field with F⊆L⊆K and let m′⁢(x) be the minimal polynomial of α over L. Also, let L′ be the field generated by the coefficients of the polynomialm′⁢(x). Thus, the minimal polynomial of α over L′ is still m′⁢(x) and L′⊆L. By the properties of the minimal polynomial, and since m⁢(α)=0, we have a divisibilitym′(x)|m(x), and so:

[K:L]=deg(m′(x))=[K:L′].

Since we know that L′⊆L, this implies that L′=L. Thus, this shows that each intermediary subfieldF⊆L⊆K corresponds with the field of definition of a (monic) factor of m⁢(x). Since the polynomial m⁢(x) has only finitely many monic factors, we conclude that there can be only finitely many subfields of K containing F.

Now suppose conversely that there are only finitely many such intermediary fields L. If F is a finite field, then so is K, and we have an explicit description of all such possibilities; all such extensions are generated by a single element. So assume F (and therefore K) are infinite. Let α1,α2,…,αn be a basis for K over F. Then K=F⁢(α1,…,αn). So if we can show that any field extension generated by two elements is also generated by one element, we will be done: simply apply the result to the last two elements αj-1 and αj repeatedly until only one is left.

So assume K=F⁢(β,γ). Consider the set of elements {β+a⁢γ} for a∈F×. By assumption, this set is infinite, but there are only finitely many fields intermediate between K and F; so two values must generate the same extension L of F, say β+a⁢γ and β+b⁢γ. This field L contains